Percentage chance of each outcome

• Jul 22nd 2007, 11:24 AM
Ken22
Percentage chance of each outcome
Alright this is more of a business problem that I'm encoutering and needed a formula for like yeterday (i.e. urgent)...so I'm just going to simplify it and knowing how to do this I'll know how to do more complex iterations.

A man is playing 5 chess matches and he has X percentage chance to win each game. We know X for each game. Against opponent 1, he has a 25% (.25) chance of winning. Against opponent 2, he has a 45% (.45) chance of winning. Against opponent 3, he has a 55% (.55) chance of winning. Against opponent 4, he has a 75% (.75) chance of winning. And against opponent 5, he has a 95% (.95) chance of winning.

What are his chances of getting 5 wins, 4 wins, 3 wins, 2 wins, 1 win, 0 win? 5 and 0 are easy enough -- .25 * .45 * .55 * .75 * .95 for 5 wins (4.4%) and those numbers subtracted from 1 multiplied by each other for 0 wins (.23%). But I need to know how to do the rest. Who he beats is irrelevent...just need to know the percentage chance of each outcome.

I have no idea if this is a complex or simple problem, but trying to Google how to do this has proven to be fruitless :(
• Jul 22nd 2007, 12:28 PM
galactus
I assume he's playing each palyer one time?. If he were playing, say, player 1 we could just use a binomial to find the probability of winning n times.

The prob. of winning 1 game if each of 5 players is played once out of 5 games would be

(.25)(.55)(.45)(.25)(.05)+(.75)(.45)(.45)(.25)(.05 )+(.75)(.55)(.55)(.25)(.05)+(.75)(.55)(.45)(.75)(. 05)+(.75)(.55)(.45)(.25)(.95)=0.056

We choose one from each player as a winner and the rest as losers then add them up.

That's one way to look at it. For 2 and 3 wins, you'll have 10 arrangements to list out. That's too many to list.
• Jul 22nd 2007, 04:01 PM
Ken22
Thank you for the reply! Is there any formula for this to make it easier?

As the sample sizes (games played) grow larger, this becomes virtually impossible (tedious to the point of impossible), even with something like Excel.
• Jul 23rd 2007, 03:35 AM
galactus
This may be a rather complicated topic, but then again, it may be something easy I am overlooking. This is a good one for the members who are well-versed in probability.

Just by adding up the outcomes, 10 in all, for 2 wins I get 25.5% chance of winning 2 games. I hope that's correct.
It would be easy to make a mistake.

Chance of winning 4 games would be (.25)(.45)(.55)(.75)(.05)+(.75)(.45)(.55)(.75)(.95 )+(.55)(.75)(.95)(.25)(.55)+(.75)(.95)(.25)(.45)(. 45)+(.95)(.25)(.45)(.55)(.25)=0.239

That seems rather high considering 5 wins is 0.044